Let A $\in M_{(n-1)},n(\mathbb{R})$ and for each $1\leq j \leq n$,let $A_j$ the matrix obtained from A by removing the j-th column. Show that:

$det (AA^t)= \sum\limits_{j=1}^n det(A_j)^2$

My first thought was using Laplace expansion,but this exercise is in the section of symmetric and exterior powers of my book. So I think maybe I should use the fact that the determinant is an alternating form . Any hints on how to proceed?


Consider $\det\pmatrix{x^t\\A}$, where $x\in \mathbb{R}^n$.

1) Notice that $\pmatrix{r_j^t\\A}$ has two equal rows for every row $r_j^t$ of $A$. Thus, $\det\pmatrix{r_j^t\\A}=0$ .

2) Expanding $\det\pmatrix{x^t\\A}$ using Laplace on the first row, we get $\det\pmatrix{x^t\\A}=x^tv,$ where $$v^t=((-1)^{1+1}\det(A_1),\ldots,(-1)^{1+n}\det(A_n)).$$

By item 1) and 2), $r_j^tv=0$ for every row $r_j$ of $A$. Therefore, $\pmatrix{v^t\\A}\pmatrix{v^t\\A}^t=\pmatrix{v^tv & 0\\0 & AA^t}$.

Therefore, $$v^tv\ \det(AA^t)=\det\left(\pmatrix{v^t\\A}\pmatrix{v^t\\A}^t\right)=\det\left(\pmatrix{v^t\\A}\right)^2=(v^tv)^2.$$

If $v\neq 0$ then $\det(AA^t)=v^tv=\sum_{i=1}^n\det(A_i)^2$.

If $v=0$ then let $y^t\neq 0$ be any row vector in the orthogonal complement of $span\{r_1,\ldots,r_{n-1}\}$.

So $\pmatrix{y^t\\A}\pmatrix{y^t\\A}^t=\pmatrix{y^ty & 0\\0 & AA^t}$ and

$$y^ty\ \det(AA^t)=\det\left(\pmatrix{y^t\\A}\pmatrix{y^t\\A}^t\right)=\det\left(\pmatrix{y^t\\A}\right)^2=(y^tv)^2=0\ (\text{since } v=0).$$

Since $y\neq 0$ then $y^ty\neq 0$ and $\det(AA^t)=0=v^tv=\sum_{i=1}^n\det(A_i)^2$.

In both cases, $v=0$ or $v\neq 0$, we get $\det(AA^t)=\sum_{i=1}^n\det(A_i)^2$.


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